Bulletin of the Iranian Mathematical Society
Vol. 27, No. 1, pp 1-18 (2001)
GENERATING THE INTEGER NULL SPACE AND
CONDITIONS FOR DETERMINATION OF AN
INTEGER BASIS USING THE ABS ALGORITHMS
H. Esmaeili, N. Mahdavi-Amiri and E. Spedicato
Department of Mathematical Sciences, Sharif University of Technology,
Tehran, Iran
Department of Mathematics, Statistics and Computer Science, University of
Bergamo, Bergamo, Italy
Abstract: We have presented a method, based on the ABS
class of algorithms, for solving the linear systems of Diophantine equations. The method provides the general solution of the
system by computing an integer solution along with an integer
matrix (generally rank decient), named as the Abaan, the integer row combinations of which generate the integer null space
of the coecient matrix. Here we show that, in general, one can
not expect that any full set of linearly independent rows of the
Abaan form an integer basis for the integer null space. We
determine the necessary and sucient conditions under which a
full rank Abaan would serve as an integer basis.
0 2000 MSC: 15A03, 15A06, 65F05, 65F30
0 Keywords: ABS algorithms, Diophantine
equation, Integer null space.
2
H. Esmaeili, N. Mahdavi-Amiri and E. Spedicato
1. Introduction
Suppose Zrepresents all integers. Consider the Diophantine linear system of equations
Ax = b x 2 Zn
(1)
where A 2 Zmn b 2 Zm and m n. By solving the system (1), rstly
we mean the determination of the existence of the solution. Secondly, if
the system has a solution , then we mean the computation of an integer
solution x
and an integer matrix H so that the rows of H , not necessarily
independent, generate the integer null space of A. That is,
Integer Null(A) = Integer Range(H T ):
Having this, the integer solutions for (1) are determined by
x = x
+ H T y
for integer vectors y . If the dimension of null space of A is r, and
H = h1 . . . hi . . . hr ]T with hi 's being linearly independent, then H T is said to be an integer
basis matrix for the integer null space of A.
Several methods, based on computing the Hermite normal form, have
been introduced before (3,5]). Recently, ABS methods have been used
extensively for solving general linear systems. In 7], we have presented
a method, based on the ABS class of algorithms, for solving the system
(1). These methods produce an integer solution x
, if it exists, and an
integer matrix, named Abaan, whose integer row combinations span
the integer null space of the coecient matrix A hence the general
integer solution of the system is readily at hand.
Section 2 explains the class of ABS methods and provides some of
its properties. In section 3, we briey discuss our algorithm for solving
the Diophantine equations (1). In this section, we then show that, in
general, one can not expect that any full set of linearly independent
Generating the integer null space and .....
3
rows of the Abaan matrix form a basis for the integer null space of
the coecient matrix. In section 4, we present necessary and sucient
conditions on the Abaan for the existence and hence the determination
of an integer basis.
2. ABS Algorithms
ABS methods have been developed by Abay, Broyden and Spedicato
1]. Consider the system of linear equations
Ax = b
(2)
where A 2 Rmn, b 2 Rm and rank(A) = m. Let A = (a1 . . . am )T ,
ai 2 Rn, i = 1 . . . m and b = (b1 . . . bm)T . Also let Ai = (a1 . . . ai)
and b(i) = (b1 . . . bi)T .
Assume x1 2 Rn arbitrary and H1 2 Rnn, Spedicato's parameter,
arbitrary and nonsingular. Note that for any x 2 Rn we can write
x = x1 + H1T q for some q 2 Rn.
The ABS class of methods are of the direct iteration types of methods for computing the general solution of (2). In the beginning of the
ith iteration, i 1, the general solution of the rst i ; 1 equation is at
hand. We realize that if xi is a solution for the rst i ; 1 equations and
if Hi 2 Rnn, with rank(Hi ) = n ; i + 1, is so that the columns of HiT
span the null space of ATi;1 , then
x = xi + HiT q
with arbitrary q 2 Rn, forms the general solution of the rst i ; 1
equations. That is, with
HiAi;1 = 0
we have
ATi;1x = b(i;1):
Now, since rank(Hi) = n ; i + 1 and HiT is a spanning matrix for
null(ATi;1), by assumption (one that is trivially valid for i = 1), then if
we let
pi = HiT zi 4
H. Esmaeili, N. Mahdavi-Amiri and E. Spedicato
with arbitrary zi 2 Rn, Broyden's parameter, then ATi;1 pi = 0 and
x() = xi ; pi for any scalar , solves the rst i ; 1 equations. We can set = i so
that xi+1 = x(i) solves the ith equation as well. If we let
T
i = ai axTi p; bi with assumption aTi pi 6= 0, then
i i
xi+1 = xi ; i pi
is a solution for the rst i equations. Now, to complete the ABS step,
Hi must be updated to Hi+1 so that Hi+1 Ai = 0. It will suce to let
Hi+1 = Hi ; uiviT
(3)
and select ui , vi so that Hi+1 aj = 0, j = 1 . . . i. The updating formula
(3) for Hi is a rank-one correction to Hi. The matrix Hi is generally
known as the Abaan. The ABS methods usually use ui = Hiai and
vi = HiT wi=wiT Hi ai, where wi , Abay's parameter, is an arbitrary vector
satisfying
wiT Hiai 6= 0:
Thus, the updating formula can be written as below:
wi Hi :
Hi+1 = Hi ; HwiaTiH
a
T
i
i i
We can now give the general steps of an ABS algorithm 1,2]. In
the algorithm below, ri+1 denotes the rank of Ai and hence the rank of
Hi+1 equals n ; ri+1.
ABS Algorithm for Solving General Linear Systems
(1) Choose x1 2 Rn, arbitrary, and H1 2 Rnn, arbitrary and nonsingular. Let i = 1, r1 = 0.
Generating the integer null space and .....
5
(2) Compute ti = aTi xi ; bi and si = Hi ai .
(3) If (si = 0 and ti = 0) then let xi+1 = xi , Hi+1 = Hi , ri+1 = ri
and go to step (7) (the ith equation is redundant). If (si = 0
and ti 6= 0) then Stop (the ith equation and hence the system
is incompatible).
(4) fsi 6= 0g Compute the search direction pi = HiT zi , where
zi 2 Rn is an arbitrary vector satisfying ziT Hiai = ziT si 6= 0.
Compute
i = ti=aTi pi
and let
xi+1 = xi ; ipi :
(5) fUpdating Hi g Update Hi to Hi+1 by
wi Hi
Hi+1 = Hi ; Hwi aTiH
a
T
i
i i
where wi 2 Rn is an arbitrary vector satisfying wiT si 6= 0.
(6) Let ri+1 = ri + 1.
(7) If i = m then Stop (xm+1 is a solution) else let i = i + 1 and
go to step (2).
We note that after the completion of the algorithm, the general solution of (2), if compatible, is written as x = xm+1 + HmT +1 q , where q 2 Rn
is arbitrary.
Below, we list certain properties of the ABS methods 2]. For simplicity, we assume rank(Ai) = i.
Hiai 6= 0 if and only if ai is linearly independent of a1 . . . ai;1.
Every row of Hi+1 corresponding to a nonzero component of wi
is linearly dependent on other rows.
The direction searches p1 . . . pi are linearly independent.
6
H. Esmaeili, N. Mahdavi-Amiri and E. Spedicato
If Li = ATi Pi, where Pi = (p1 . . . pi), then Li is a nonsingular
lower triangular matrix.
The set of directions p1 . . . pi together with independent columns
of HiT+1 form a basis for Rn.
The matrix Wi = (w1 . . . wi) has full column rank and Null(HiT+1) =
Range(Wi), while Null(Hi+1) = Range(Ai).
If rows j1 . . . ji of Wi are linearly independent then the same
rows of Hi+1 are linearly dependent and vice versa. Specially,
each row of Hi+1 corresponding to a nonzero element of wi is
dependent.
If si 6= 0, then rank(Hi+1) = rank(Hi) ; 1.
The updating formula Hi can be written as:
Hi+1 = H1 ; H1Ai (WiT H1Ai );1WiT H1
where WiT H1Ai is strongly nonsingular (the determinants of all
of its main principal submatrices are nonzero).
3. Solving Linear Diophantine Equations
Consider the linear Diophantine system of equations
Ax = b x 2 Zn
(4)
where A 2 Zmn, b 2 Zm. The following results indicate how to choose
H1, zi and wi within the ABS algorithms to obtain the integer solution
of (4) see 7]. Assume i to be the greatest common divisor (gcd) of the
components of Hi ai.
Theorem 1. Let A be full rank and suppose that the Diophantine
system (4) is solvable. Consider the sequence of Abaans generated by
the basic ABS algorithm with the following parameter choices:
(a) H1 is unimodular (an integer matrix whose inverse is also integer with the modules of its determinant equal to 1).
Generating the integer null space and .....
7
(b) For i = 1 . . . m, the integer vector wi is such that wiT Hiai =
i i = gcd(Hiai).
Then the following properties are true:
(c) The sequence of Abaans generated by the algorithm is welldened and consists of integer matrices.
(d) If xi+1 is a special integer solution of the rst i equations, then
any integer solution x of the rst i equations can be written in
the form x = xi+1 + HiT+1 q for some integer vector q .
Theorem 2. Let A be full rank and consider the sequence of matrices Hi generated by the basic ABS algorithm with parameter choices as in
Theorem 1. Let the initial point x1 in the basic ABS algorithm be an arbitrary integer vector and let zi be chosen such that ziT Hiai = gcd(Hiai ).
Then system (4) has integer solutions if and only if gcd(Hiai ) divides
aTi xi ; bi for i = 1 . . . m.
Note: The computation of i and solving for an integer y in sTi y = i,
where si = Hi ai , can be achieved by Rosser's algorithm 9,10].
It follows from the above theorems that if there exists a solution for
the system (4), then x = xm+1 + HmT +1 q , with arbitrary q 2 Z n , forms
the general solution of (4). We continue by an analysis showing that, in
general, any n ; m independent columns of HmT +1 would not be an integer
basis for the integer null space of the matrix A. For simplicity, let x1 = 0,
x
= xm+1 , H = Hm+1 and assume rank(A) = m. Let H
2 Z(n;m)m be
a matrix composed of any set of n ; m linearly independent rows of H .
We show that, in general, it can not be expected that
x = x
+ H
T y y 2 Zn;m
(5)
provide all the integer solutions for (4).
Let P = (p1 . . . pm) be the matrix of search directions obtained
from the application of an ABS algorithm in solving the system (4).
Let K = (P H
T ). We know that the matrix K is nonsingular and
AK = A(P H
T ) = (AP AH
T ) = (L 0)
8
H. Esmaeili, N. Mahdavi-Amiri and E. Spedicato
where L is lower triangular and nonsingular. The next theorem states
conditions under which x = x
+ H
T y , y 2 Zn;m, forms the general
solution of (4). We note that since x1 = 0 then we can write x
= Pq
for some vector q . Hence we have b = Ax
= APq = Lq . Since L is
nonsingular then q = L;1 b and x
= PL;1 b.
Theorem 3. The expression x = x
+ H
T y is the general solution for
the Diophantine system (4) when the matrix K = (P H
T ) is unimodular.
Proof: The vector x = x
+ H
T y for any integer vector y is integer.
For such x we have Ax = b, since A!H
T = 0. Now suppose x 2 Zn
u
satises Ax = b. Let K ;1 x = u = u1 . Since K is unimodular then u
2
in an integer vector with u1 2 Zm and u2 2 Zn;m. Now, we can write
!
u
b = Ax = AKK x = AKu = (L 0) u1 = Lu1 u1 = L;1 b:
2
;1
Hence
!
u
x = Ku = (P H
T ) u1 = Pu1 +H
T u2 = PL;1b+ H
T u2 = x
+ H
T u2 : 2
2
Note: Using the geometry of numbers, the converse of the above
theorem is also established (see 4]).
Consider the single Diophantine equation aT x = 0 x 2 Zn. Assume
H1 is unimodular. Let = gcd(s), where s = H1a, and assume z is so
that sT z = . We know from Rosser's algorithm that the rst component
of s has the largest magnitude in s and is nonzero, since s 6= 0. Thus
k s k1= jsT e1j 6= 0, where e1 is the rst column of the identity matrix.
Suppose p = H1T z is the search direction of the ABS method for solving
aT x = 0, and w is an integer vector so that wT e1 6= 0 and sT w = .
Then, from the ABS properties, the rst row of H2 is dependent and
we can dene H
to represent the independent rows of H2 by
!
!
T H1
swT H =
E
I
;
H
= E H1 ; Hw1aw
1
T H1a
Generating the integer null space and .....
9
where E is the identity matrix with its rst row deleted. From Theorem
3, the vector x = H
T y , where y is an arbitrary integer vector, is the
general solution for aT x = 0, x 2 Zn, when M = (p H
T ) is unimodular.
Since H1 is unimodular, then
M = (p H
T ) = H1T z H1T I ; ws
T
! !
is unimodular if and only if the matrix
ET
= HT
1
! !
T
z I ; ws E T
! !
T
z I ; ws E T
is unimodular. Let
where
K = (z B )
!
T
B = I ; ws E T :
We note that K is nonsingular. We shall make use of the determinant
of K in subsequent discussions. Note the following lemma.
Lemma 1. If K = (z B), where B = (I ; wsT =)E T and E is
obtained from the identity matrix with its rst row deleted, then det K =
wT e1 .
Proof: The matrix K = (z B) is a rank one correction to the matrix
1 B ), since
K
= I ; wsT = = (Ke
1 )eT1 :
K = K
+ (z ; Ke
Note that
where
T
T
K = I ; e1eT1 ; ws + e1 s weT1 + zeT1
= K1 + (z ; e1 )eT1 K1 = I + wv T 10
and
Hence
H. Esmaeili, N. Mahdavi-Amiri and E. Spedicato
vT = 1 (eT1 s)eT1 ; sT ]:
det K1 = 1 + v T w = (e1 s)( e1 w) 6= 0:
T
T
Thus K1 is always properly dened and always nonsingular and we may
write K = K1K2, where
K2 = I + K1;1(z ; e1 )eT1 :
Therefore
det K2 = 1 + eT1 K1;1(z ; e1 ):
Now, expanding K1;1 by the Sherman-Morrison-Woodbury formula 8]
gives det K2 = =(eT1 s). Since det K = detK1 det K2, the result follows. 2
Therefore, K and hence M are unimodular if and only if the integer
vector w satises wT e1 = 1 and wT s = , conditions not expected to
hold in general. Egervary's method 6] is a special ABS method with
the selections H1 = I , x1 = 0 and w = z . Since the Diophantine system
sT z = , zT e1 = 1, lacks integer solutions in general, then Egervary's
claim that any set of independent columns of H T provides an integer
basis for the general integer solutions is refuted. The next example
validates this statement.
Example 1. If we use Egervary's method for solving the homogeneous Diophantine equation
aT x = 0 aT = (1 1 1)
(6)
with z = (2 2 ;3)T , we obtain:
2;1 ;2 ;23
H = H2T = I ; zaT = 64;2 ;1 ;275 :
3
There are three possible choices for H
T .
3
4
Generating the integer null space and .....
11
(a) H
T = (H2T e1 H2T e2). The vector x
= (;2 1 1)T satises (6).
The only solution for H
T t = x
is t = (;4=3 5=3)T .
(b) H
T = (H2T e1 H2T e3). The vector x
= (;2 1 1)T satises (6).
The only solution for H
T t = x
is t = (;3 5=2)T .
(c) H
T = (H2T e2 H2T e3). The vector x
= (;3 2 1)T satises (6).
The only solution for H
T t = x
is t = (5 ;7=2)T .
We see that in all the possible three cases there is at least one integer
solution x
for (6) not being generated by an integer combinations of
columns of H
T .
Note: For the homogeneous Diophantine system (case b = 0 in (4)),
Egervary 6] presented a method being now a special version of the ABS
algorithms with H1 = I , x1 = 0, zi = wi for all i. We realize that the
general solution in this case is written as x = HmT +1 y , where y 2 Zn
is arbitrary. Egervary believed that with r being the rank of A, any
set of n ; r independent columns of HmT +1 would form an integer basis
for the integer solutions of the system. The results given above clearly
invalidates this belief (see also 7]).
In the next section, we introduce the necessary and sucient conditions for producing an integer basis from the Abaan matrix. There,
we return to Example 1 again and show how to determine an integer
basis using these conditions.
4. The Necessary and Sucient Conditions
Assume rank(A) = m. We now determine conditions under which one
can eliminate m columns of HmT +1 and obtain an integer basis, composed
of n ; m linearly independent columns, for Null(A) \ Zn. For convenience, let H = Hm+1 . Let W = (w1 . . . wm) 2 Znm be the matrix
with Abaan parameters as its columns. We know that rank(W ) = m.
According to ABS properties, the rows of H corresponding to m linearly
independent rows of W are linearly dependent. Since rank(H ) = n ; m
12
H. Esmaeili, N. Mahdavi-Amiri and E. Spedicato
!
H
then, without loss of generality, we can write H = U H
, where
H
2 Z(n;m)n corresponds to the n ; m linearly independent rows of H
and U 2 Rm(n;m). We can now let W T = (V T T T ), where T 2 Zmm
is nonsingular. Since
0 = H T W = H
T V + H
T U T T
then H
T U T = ;H
T V T ;1 and whereof
U T = ;V T ;1:
We emphasize that U is not necessarily an integer matrix. Fix an arbitrary vector y 2 Null(A) \ Zn. The full column rank system
H
T t = y
(7)
has a unique solution. The following lemma gives the correspondence
between t, the unique solution of (7), and the solutions of the system
H T x = y:
Lemma 2. x is a solution of (8) if and only if we have
! T!
t
;U q
x=
+
0
Im
with t being the unique solution of (7) and q 2 Rm.
!
Proof: Let x = xxn;mm be a solution of (8). Then
y = H T x = H
T xn;m + H
T U T xm = H
T (xn;m + U T xm ):
Since (7) has a unique solution then t = xn;m + U T xm and hence
t ! ;U T !
x = 0 + I xm :
m
(8)
Generating the integer null space and .....
13
Conversely, let t be the unique solution of (7) and q 2 Rm. Consider
! !
;
UT
x = 0 + I q:
m
t
We have
H T x = H
T t ; H
T U T q + H
T U T q = H
T t = y:
Therefor, x is a solution of (8). 2
We saw before that for any y 2 Null(A) \ Zn, the integer vector
x = H1;T y solves (8). Let H1;1 = (H11T H21T ). H1 being unimodular,
both H11 and H21 are integer matrices. Applying Lemma 2, for some
q 2 Rm and t, the unique solution of (7), we must have:
!
! !
H11
t
;
UT
x = H y = H y = 0 + I q:
21
m
;T
1
Hence we have:
H11y = t ; U T q
H21y = q:
Now, for x = H1;T y since y is an integer vector then both q and t ; U T q
must be integer vectors. Therefore, to have t integer it would suce
that H
T be constructed from H T in such a way that the corresponding
matrix U be integer (or can be reduced to an integer matrix). On the
other hand, we realize that no column of H
T should have a common
divisor other than one (that is, the greatest common divisor for every
column should be one), since the system H
T t = y will have noninteger
solutions, otherwise. Having this in mind, we consider reducing the
matrix H T = (H
T H
T U T ) accordingly. To make the columns of H
T
be relatively prime, we multiply H T by D on the right, where D is a
diagonal matrix as below
D
0 !
D= 0 I m
14
H. Esmaeili, N. Mahdavi-Amiri and E. Spedicato
with D
ii = 1=gcd(H T ei ). Therefore, we have
H~ T = H T D = (H^ T H^ T U^ T )
where
U^ T = D
;1 U T :
H^ T = H
T D
Now, let adj (T ) be the classical adjoint of T (that is, T ;1 = adj (T )=det T ).
Since U T = ;V T ;1 , we can write
U^ T = ;D
;1 V T ;1 = ;D
;1V adj (T )=det T:
The following theorem states the necessary and sucient conditions for
= y to be integer.
the solution of the system H^ T t = H
T Dt
Theorem 4. Let y 2 Null(A) \ Zn be arbitrary. The solution t^ for
the full column rank system H^ T t = y is an integer vector if and only if
det T j D
;1V adj (T ). (ajb means a divided by b is an integer.)
Proof: We saw that x = H1;T y 2 Zn, for any y 2 Null(A) \ Zn,
satises H T x = y . Thus, for x~ = D;1 H1;T y 2 Zn we have H~ T x~ = y .
Let x~ = (~xTn;m x~Tm)T and suppose that det T jD
;1V adj (T ). Then U^ is
an integer matrix and
!
;1 ! ! ;1 !
x~n;m
D
0 H11y
D H11y
;1 ;T
x~ = x~
= D H1 y = 0 I
=
H21y
H21y :
m
m
Thus, the vector
^t = x~n;m + U^ T x~m = D
;1 H11y + D
;1 U T H21y = D
;1 (H11y + U T H21y )
is integer and
!
H y
H^ T ^t = H
T D
D
;1 (H11y + U T H21y ) = H
T (In;m U T ) H11y
21
T
T
T
;T
T
= (H H U )H1 y = H x = y:
Conversely, suppose that for any y 2 Null(A) \ Zn, the solution t^ for
H^ T t = y be integer. Applying Lemma 2, the integer solutions for H~ T x =
y can be written as
^! ^ T !
t
x~ = 0 + ;IU q
m
Generating the integer null space and .....
15
where q 2 Zm. Since H~ T x~ = y , then
t^ = x~n;m + U^ T x~m = D
;1H11y + D
;1U T H21y
H y!
;1
;1
T
= (D
D
U ) H11y = (D
;1 D
;1U T )H1;T y:
21
Note that, for any y 2 Null(A) \ Zn, H1;T y , ^t and D;1 are integers.
Therefore, D
;1U T must also be an integer matrix because the rows
of H21 are relatively prime. From D
;1U T = ;D
;1 V adj (T )=det T , it
follows that det T j D
;1V adj (T ). 2
We now return to case (b) in Example 1. We have,
2;1 ;23
021
H
T = 64;2 ;275 W = B
@ 2 CA T = (2)
3 4
;3
!
!
2
1
0
;
1
V = ;3 U = 2 (2 ;3) D
= 0 1=2 :
We see that
2;1 ;13
H^ T = 64;2 ;175 3
2
and
U^ = (;1 3)
an integer vector now. The solution for H^ T t = y , with y = (;2 1 1)T , is
the integer vector t^ = (;3 5)T . On the other hand, any y 2 Null(aT ) \
Z3 can be written as y = (; ; )T , where and are arbitrary
integers. For any such y , the solution for H^ T t = y is given by t^ =
(;2 ; 3 + 2 )T . Therefore, in this case, we have
fx 2 Z3 j aT x = 0g = fH^ T q j q 2 Z2g:
Similar developments for case (c) will also result in an integer basis for
Null(aT ) \ Z3. At the same time, we note that no integer basis can be
16
H. Esmaeili, N. Mahdavi-Amiri and E. Spedicato
obtained from the matrix in case (a). Therefore, we observe that an
integer basis can not necessarily be obtained from any set of linearly
independent columns of H T .
Determining an Integer Basis
Considering the ABS properties, instead of deleting the m dependent
columns of H T all at the same time, deletions can be made in steps.
From the ABS properties, at the end of the i-th iteration, an independent column of HiT+1 can be identied and subsequently deleted. We
know that any column of HiT+1 corresponding to a nonzero component of
wi is linearly dependent on the other columns. Let wi = (w1i . . . wni)T
and suppose wki 6= 0, for some k, 1 k n. Then T = wki,
V = (w1i . . . wk;1i wk+1i . . . wni)T and U T = ;V=wki. We can now
state the following rule for the deletion of a dependent column of HiT+1
at the end of the i-th iteration.
Deletion Rule For a Dependent Column
Let j = gcd(HiT+1ej ) for all j . Delete the k-th column of HiT+1 , where
wiT ek 6= 0 and wiT ek jj wiT ej for all j .
We note that one can not expect the satisfaction of the above conditions in all cases. Thus, the ABS approach may signal the failure by
recognizing that an integer basis may not be obtained. Nevertheless,
the columns of H T span Null(A) \ Zn and, as such, the general integer
solutions may be obtained using H . The following example illustrates
the point.
Example 2. Consider the Diophantine system below:
aT x = 0 aT = (1 3 ;2):
With the choice z = (2 3 5)T , we have:
2;1 ;6 4 3
H T = I ; zaT = 64;3 ;8 6 75 :
;5 ;15 11
(9)
Generating the integer null space and .....
17
We see that every column of H T is relatively prime. Consider the system
H
T t = y , where y = (;1 1 1)T is a solution of (9).
(a) By selecting H
T = (H T e1 H T e2 ), we have t = (;7=5 2=5)T .
(b) By selecting H
T = (H T e1 H T e3 ), we have t = (;5=3 ;2=3)T .
(c) By selecting H
T = (H T e2 H T e3 ), we have t = (5=2 7=2)T .
We see that in all the possible three cases there is at least one integer solution for (9) not being generated by an integer combinations of columns
of H
T .
5. Conclusions
We saw how an integer Abaan (not necessarily full rank) matrix is
obtained by use of the ABS methods for solving a linear Diophantine
system of equations. The integer combinations of the rows of the Abafan span the integer null space of the coecient matrix. We proved that,
in general, it can not be expected that the resulting Abaan would contain an integer basis for this integer null space. Finally, we specied
the necessary and sucient conditions under which the Abaan would
present an integer basis.
Acknowledgements. The work of the rst two authors has been
supported by the Research Council of Sharif University of Technology.
The authors are grateful to the referees for their constructive comments,
specially to one anonymous referee who provided a more concise proof
of Lemma 1.
References
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equations, Numer. Math., 45 (1984) 361-376.
18
H. Esmaeili, N. Mahdavi-Amiri and E. Spedicato
2] J. Abay, E. Spedicato, ABS Projection Algorithms: Mathematical Techniques for Linear and Nonlinear Equations, Ellis Horwood, Chichester,
1989.
3] G.H. Bradly, Algorithms for Hermite and Smith normal matrices and
linear Diophantine equations, Math. Comp., 25 (1971) 897-907.
4] J.W.S. Cassels, An Introduction to the Geometry of Numbers, Springer,
Berlin, 1959.
5] T.J. Chou, E.E. Collins, Algorithms for the solution of systems of linear
Diophantine equations, SIAM J. Comp., 11 (1982) 786-708.
6] E. Egervary, On rank-diminishing operations and their applications to the
solution of linear equations, ZAMP, 9 (1960) 376-386.
7] H. Esmaeili, N. Mahdavi-Amiri, E. Spedicato, A class of ABS algorithms
for Diophantine linear systems, to appear in Numericshe Mathematik.
8] G.H. Golub, C.F. Van Loan, Matrix Computations, Johns Hopkins University Press, 1989.
9] S. Morito, H.M. Salkin, Using the Blankinship algorithm to nd the general solution of a linear Diophantine equation, Acta Inf., 13 (1980) 379382.
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Monthly, 48 (1941) 662-666.
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